đk: \(\hept{\begin{cases}x^2-2x+5\ge0\\4x+5\ge0\end{cases}}\Leftrightarrow x\ge\frac{-5}{4}\)

Ta có: \(x^3-2x^2-\sqrt{x^2-2x+5}=2\sqrt{4x+5}-5x-4\)

\(\Leftrightarrow3x^3-6x^2+15x+12-3\sqrt{x^2-2x+5}-6\sqrt{4x+5}=0\)

\(\Leftrightarrow3\left(x+1-\sqrt{x^2-2x+5}\right)+2\sqrt{4x+5}\left(\sqrt{4x+5}-3\right)+3x^3-6x^2+4x-1=0\)

\(\Leftrightarrow\frac{12\left(x-1\right)}{x+1+\sqrt{x^2-2x+5}}+\frac{8\left(x-1\right)\sqrt{4x+5}}{\sqrt{4x+5}+3}+\left(x-1\right)\left(3x^2-3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{12}{x+1+\sqrt{x^2-2x+5}}+\frac{8\sqrt{4x+5}}{\sqrt{4x+5}+3}+3x^2-3x+1\right)=0\Leftrightarrow x=1\)

Từ pt ta có: \(-\left(1+x^4\right)=\text{ax}^3+bx^2+cx\)

Áp dụng BĐT B.C.S:

\(\left(1+x^4\right)^2=\left(\text{ax}^3+bx^2+cx\right)^2\le\left(a^2+b^2+c^2\right)\left(x^6+x^4+x^2\right)\)\(\Rightarrow\left(a^2+b^2+c^2\right)\ge\frac{\left(1+x^4\right)^2}{x^6+x^4+x^2}\left(1\right)\)

Mặt khác: \(\frac{\left(1+x^4\right)^2}{x^6+x^4+x^2}\ge\frac{4}{3}\left(2\right)\)

Thật vậy: \(\left(2\right)\Leftrightarrow3\left(1+2x^4+x^8\right)\ge4\left(x^6+x^4+x^2\right)\)

\(\Leftrightarrow3x^8-4x^6+2x^4-4x^2+3\ge0\)

\(\Leftrightarrow\left(x^2-1\right)^2\left(3x^4+2x^2+3\right)\ge0\)(luôn đúng)

Từ 1 và 2 : \(a^2+b^2+c^2\ge\frac{4}{3}\)

Dấu '=' xảy ra khi và chỉ khi \(\orbr{\begin{cases}a=b=c=\frac{2}{3}\left(x=1\right)\\a=b=c=\frac{-2}{3}\left(x=-1\right)\end{cases}}\)

HD: \(\overrightarrow{BC}=\frac{-2}{3}\overrightarrow{AM}+\frac{4}{3}\overrightarrow{AN};\overrightarrow{CD}=\frac{-4}{3}\overrightarrow{AM}+\frac{2}{3}\overrightarrow{AN}\)

Bạn xem hình nha

2 hình nha hok tốt

\(a)13-\left(40-X\right)=35\)

\(\Leftrightarrow40-X=13-35\)

\(\Leftrightarrow40-X=-22\)

\(\Leftrightarrow X=40-\left(-22\right)\)

\(\Leftrightarrow X=62\)

Vậy \(X=62\)

\(b)14-3x\left(5-X\right)=8\)

\(\Leftrightarrow3x\left(5-X\right)=14-8\)

\(\Leftrightarrow3x\left(5-X\right)=6\)

\(\Leftrightarrow5-X=6:3\)

\(\Leftrightarrow5-X=3\)

\(\Leftrightarrow X=5-3\)

\(\Leftrightarrow X=2\)

Vậy \(X=2\)

\(c)\left(3X-2\right)x\left(5+X\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3X-2=0\\5+X=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3X=0+2\\X=0-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3X=2\\X=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}X=2:3\\X=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}X=\frac{2}{3}\\X=-5\end{cases}}\)

Vậy \(X\in\left\{-5;\frac{2}{3}\right\}\)

\(d)\left(3X-6\right)x3=3^4\)

\(\Leftrightarrow\left(3X-6\right)x3=81\)

\(\Leftrightarrow3X-6=81:3\)

\(\Leftrightarrow3X-6=27\)

\(\Leftrightarrow3X=27+6\)

\(\Leftrightarrow3X=33\)

\(\Leftrightarrow X=33:3\)

\(\Leftrightarrow X=11\)

Vậy\(X=11\)

\(\left|\overrightarrow{MA}+\overrightarrow{MC}-\overrightarrow{MN}\right|=\left|\overrightarrow{MA}+\overrightarrow{MD}+\overrightarrow{DC}-\overrightarrow{MN}\right|\)\(=\left|\overrightarrow{DC}-\frac{1}{2}\overrightarrow{DC}-\frac{1}{2}\overrightarrow{AB}\right|=\left|\overrightarrow{DC}-\frac{3}{4}\overrightarrow{DC}\right|=\frac{1}{A}DC=\frac{a}{2}\)

mk bận đi ch nên chỉ tạm câu a nha 

vẽ 3 đường trung tuyến AD ; BE ; CF 

VT = 

\(GA+GB+GC\)   ( nhớ thêm dấu vec tơ nha ) 

\(=-\frac{2}{3}AD-\frac{2}{3}BE-\frac{2}{3}CF\)  

\(=-\frac{2}{3}\cdot\frac{1}{2}\left(AB+BC\right)-\frac{2}{3}\cdot\frac{1}{2}\left(BA+BC\right)-\frac{2}{3}\cdot\frac{1}{2}\left(CA+CB\right)\)     ( quy tắc hình bình hành ) 

\(=-\frac{1}{3}\left(AB+AC\right)-\frac{1}{3}\left(BA+BC\right)-\frac{1}{3}\left(CA+CB\right)\) 

\(=-\frac{1}{3}AB-\frac{1}{3}AC-\frac{1}{3}BA-\frac{1}{3}BC-\frac{1}{3}CA-\frac{1}{3}CB\)    

\(=0=VP\)

.... chua hoc

Giả sử \(x\ge y\ge z\)cũng được mà.